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    <p><span id="tut-fp-issues"></span></p>
    <h1 id="floating-point-arithmetic-issues-and-limitations">
      <span class="section-number">15. </span>Floating Point Arithmetic: Issues
      and Limitations<a
        href="#floating-point-arithmetic-issues-and-limitations"
        class="headerlink"
        title="Permalink to this headline"
        >¶</a
      >
    </h1>
    <p>
      Floating-point numbers are represented in computer hardware as base 2
      (binary) fractions. For example, the decimal fraction
    </p>
    <pre><code>0.125</code></pre>
    <p>
      has value 1/10 + 2/100 + 5/1000, and in the same way the binary fraction
    </p>
    <pre><code>0.001</code></pre>
    <p>
      has value 0/2 + 0/4 + 1/8. These two fractions have identical values, the
      only real difference being that the first is written in base 10 fractional
      notation, and the second in base 2.
    </p>
    <p>
      Unfortunately, most decimal fractions cannot be represented exactly as
      binary fractions. A consequence is that, in general, the decimal
      floating-point numbers you enter are only approximated by the binary
      floating-point numbers actually stored in the machine.
    </p>
    <p>
      The problem is easier to understand at first in base 10. Consider the
      fraction 1/3. You can approximate that as a base 10 fraction:
    </p>
    <pre><code>0.3</code></pre>
    <p>or, better,</p>
    <pre><code>0.33</code></pre>
    <p>or, better,</p>
    <pre><code>0.333</code></pre>
    <p>
      and so on. No matter how many digits you’re willing to write down, the
      result will never be exactly 1/3, but will be an increasingly better
      approximation of 1/3.
    </p>
    <p>
      In the same way, no matter how many base 2 digits you’re willing to use,
      the decimal value 0.1 cannot be represented exactly as a base 2 fraction.
      In base 2, 1/10 is the infinitely repeating fraction
    </p>
    <pre><code>0.0001100110011001100110011001100110011001100110011...</code></pre>
    <p>
      Stop at any finite number of bits, and you get an approximation. On most
      machines today, floats are approximated using a binary fraction with the
      numerator using the first 53 bits starting with the most significant bit
      and with the denominator as a power of two. In the case of 1/10, the
      binary fraction is <code>3602879701896397 / 2 ** 55</code> which is close
      to but not exactly equal to the true value of 1/10.
    </p>
    <p>
      Many users are not aware of the approximation because of the way values
      are displayed. Python only prints a decimal approximation to the true
      decimal value of the binary approximation stored by the machine. On most
      machines, if Python were to print the true decimal value of the binary
      approximation stored for 0.1, it would have to display
    </p>
    <pre><code>&gt;&gt;&gt; 0.1
0.1000000000000000055511151231257827021181583404541015625</code></pre>
    <p>
      That is more digits than most people find useful, so Python keeps the
      number of digits manageable by displaying a rounded value instead
    </p>
    <pre><code>&gt;&gt;&gt; 1 / 10
0.1</code></pre>
    <p>
      Just remember, even though the printed result looks like the exact value
      of 1/10, the actual stored value is the nearest representable binary
      fraction.
    </p>
    <p>
      Interestingly, there are many different decimal numbers that share the
      same nearest approximate binary fraction. For example, the numbers
      <code>0.1</code> and <code>0.10000000000000001</code> and
      <code>0.1000000000000000055511151231257827021181583404541015625</code> are
      all approximated by <code>3602879701896397 / 2 ** 55</code>. Since all of
      these decimal values share the same approximation, any one of them could
      be displayed while still preserving the invariant
      <code>eval(repr(x)) == x</code>.
    </p>
    <p>
      Historically, the Python prompt and built-in
      <a
        href="https://docs.python.org/3/library/functions.html#repr"
        class="reference internal"
        title="repr"
        ><code class="sourceCode python"
          ><span class="bu">repr</span>()</code
        ></a
      >
      function would choose the one with 17 significant digits,
      <code>0.10000000000000001</code>. Starting with Python 3.1, Python (on
      most systems) is now able to choose the shortest of these and simply
      display <code>0.1</code>.
    </p>
    <p>
      Note that this is in the very nature of binary floating-point: this is not
      a bug in Python, and it is not a bug in your code either. You’ll see the
      same kind of thing in all languages that support your hardware’s
      floating-point arithmetic (although some languages may not
      <em>display</em> the difference by default, or in all output modes).
    </p>
    <p>
      For more pleasant output, you may wish to use string formatting to produce
      a limited number of significant digits:
    </p>
    <pre><code>&gt;&gt;&gt; format(math.pi, &#39;.12g&#39;)  # give 12 significant digits
&#39;3.14159265359&#39;

&gt;&gt;&gt; format(math.pi, &#39;.2f&#39;)   # give 2 digits after the point
&#39;3.14&#39;

&gt;&gt;&gt; repr(math.pi)
&#39;3.141592653589793&#39;</code></pre>
    <p>
      It’s important to realize that this is, in a real sense, an illusion:
      you’re simply rounding the <em>display</em> of the true machine value.
    </p>
    <p>
      One illusion may beget another. For example, since 0.1 is not exactly
      1/10, summing three values of 0.1 may not yield exactly 0.3, either:
    </p>
    <pre><code>&gt;&gt;&gt; .1 + .1 + .1 == .3
False</code></pre>
    <p>
      Also, since the 0.1 cannot get any closer to the exact value of 1/10 and
      0.3 cannot get any closer to the exact value of 3/10, then pre-rounding
      with
      <a
        href="https://docs.python.org/3/library/functions.html#round"
        class="reference internal"
        title="round"
        ><code class="sourceCode python"
          ><span class="bu">round</span>()</code
        ></a
      >
      function cannot help:
    </p>
    <pre><code>&gt;&gt;&gt; round(.1, 1) + round(.1, 1) + round(.1, 1) == round(.3, 1)
False</code></pre>
    <p>
      Though the numbers cannot be made closer to their intended exact values,
      the
      <a
        href="https://docs.python.org/3/library/functions.html#round"
        class="reference internal"
        title="round"
        ><code class="sourceCode python"
          ><span class="bu">round</span>()</code
        ></a
      >
      function can be useful for post-rounding so that results with inexact
      values become comparable to one another:
    </p>
    <pre><code>&gt;&gt;&gt; round(.1 + .1 + .1, 10) == round(.3, 10)
True</code></pre>
    <p>
      Binary floating-point arithmetic holds many surprises like this. The
      problem with “0.1” is explained in precise detail below, in the
      “Representation Error” section. See
      <a href="http://www.lahey.com/float.htm" class="reference external"
        >The Perils of Floating Point</a
      >
      for a more complete account of other common surprises.
    </p>
    <p>
      As that says near the end, “there are no easy answers.” Still, don’t be
      unduly wary of floating-point! The errors in Python float operations are
      inherited from the floating-point hardware, and on most machines are on
      the order of no more than 1 part in 2**53 per operation. That’s more than
      adequate for most tasks, but you do need to keep in mind that it’s not
      decimal arithmetic and that every float operation can suffer a new
      rounding error.
    </p>
    <p>
      While pathological cases do exist, for most casual use of floating-point
      arithmetic you’ll see the result you expect in the end if you simply round
      the display of your final results to the number of decimal digits you
      expect.
      <a
        href="https://docs.python.org/3/library/stdtypes.html#str"
        class="reference internal"
        title="str"
        ><code class="sourceCode python"><span class="bu">str</span>()</code></a
      >
      usually suffices, and for finer control see the
      <a
        href="https://docs.python.org/3/library/stdtypes.html#str.format"
        class="reference internal"
        title="str.format"
        ><code class="sourceCode python"
          ><span class="bu">str</span>.<span class="bu">format</span>()</code
        ></a
      >
      method’s format specifiers in
      <a
        href="https://docs.python.org/3/library/string.html#formatstrings"
        class="reference internal"
        ><span class="std std-ref">Format String Syntax</span></a
      >.
    </p>
    <p>
      For use cases which require exact decimal representation, try using the
      <a
        href="https://docs.python.org/3/library/decimal.html#module-decimal"
        class="reference internal"
        title="decimal: Implementation of the General Decimal Arithmetic  Specification."
        ><code class="sourceCode python">decimal</code></a
      >
      module which implements decimal arithmetic suitable for accounting
      applications and high-precision applications.
    </p>
    <p>
      Another form of exact arithmetic is supported by the
      <a
        href="https://docs.python.org/3/library/fractions.html#module-fractions"
        class="reference internal"
        title="fractions: Rational numbers."
        ><code class="sourceCode python">fractions</code></a
      >
      module which implements arithmetic based on rational numbers (so the
      numbers like 1/3 can be represented exactly).
    </p>
    <p>
      If you are a heavy user of floating point operations you should take a
      look at the Numerical Python package and many other packages for
      mathematical and statistical operations supplied by the SciPy project. See
      &lt;<a href="https://scipy.org/" class="reference external"
        >https://scipy.org</a
      >&gt;.
    </p>
    <p>
      Python provides tools that may help on those rare occasions when you
      really <em>do</em> want to know the exact value of a float. The
      <a
        href="https://docs.python.org/3/library/stdtypes.html#float.as_integer_ratio"
        class="reference internal"
        title="float.as_integer_ratio"
        ><code class="sourceCode python"
          ><span class="bu">float</span>.as_integer_ratio()</code
        ></a
      >
      method expresses the value of a float as a fraction:
    </p>
    <pre><code>&gt;&gt;&gt; x = 3.14159
&gt;&gt;&gt; x.as_integer_ratio()
(3537115888337719, 1125899906842624)</code></pre>
    <p>
      Since the ratio is exact, it can be used to losslessly recreate the
      original value:
    </p>
    <pre><code>&gt;&gt;&gt; x == 3537115888337719 / 1125899906842624
True</code></pre>
    <p>
      The
      <a
        href="https://docs.python.org/3/library/stdtypes.html#float.hex"
        class="reference internal"
        title="float.hex"
        ><code class="sourceCode python"
          ><span class="bu">float</span>.<span class="bu">hex</span>()</code
        ></a
      >
      method expresses a float in hexadecimal (base 16), again giving the exact
      value stored by your computer:
    </p>
    <pre><code>&gt;&gt;&gt; x.hex()
&#39;0x1.921f9f01b866ep+1&#39;</code></pre>
    <p>
      This precise hexadecimal representation can be used to reconstruct the
      float value exactly:
    </p>
    <pre><code>&gt;&gt;&gt; x == float.fromhex(&#39;0x1.921f9f01b866ep+1&#39;)
True</code></pre>
    <p>
      Since the representation is exact, it is useful for reliably porting
      values across different versions of Python (platform independence) and
      exchanging data with other languages that support the same format (such as
      Java and C99).
    </p>
    <p>
      Another helpful tool is the
      <a
        href="https://docs.python.org/3/library/math.html#math.fsum"
        class="reference internal"
        title="math.fsum"
        ><code class="sourceCode python">math.fsum()</code></a
      >
      function which helps mitigate loss-of-precision during summation. It
      tracks “lost digits” as values are added onto a running total. That can
      make a difference in overall accuracy so that the errors do not accumulate
      to the point where they affect the final total:
    </p>
    <pre><code>&gt;&gt;&gt; sum([0.1] * 10) == 1.0
False
&gt;&gt;&gt; math.fsum([0.1] * 10) == 1.0
True</code></pre>
    <p><span id="tut-fp-error"></span></p>
    <h2 id="representation-error">
      <span class="section-number">15.1. </span>Representation Error<a
        href="#representation-error"
        class="headerlink"
        title="Permalink to this headline"
        >¶</a
      >
    </h2>
    <p>
      This section explains the “0.1” example in detail, and shows how you can
      perform an exact analysis of cases like this yourself. Basic familiarity
      with binary floating-point representation is assumed.
    </p>
    <p>
      <em>Representation error</em> refers to the fact that some (most,
      actually) decimal fractions cannot be represented exactly as binary (base
      2) fractions. This is the chief reason why Python (or Perl, C, C++, Java,
      Fortran, and many others) often won’t display the exact decimal number you
      expect.
    </p>
    <p>
      Why is that? 1/10 is not exactly representable as a binary fraction.
      Almost all machines today (November 2000) use IEEE-754 floating point
      arithmetic, and almost all platforms map Python floats to IEEE-754 “double
      precision”. 754 doubles contain 53 bits of precision, so on input the
      computer strives to convert 0.1 to the closest fraction it can of the form
      <em>J</em>/2**<em>N</em> where <em>J</em> is an integer containing exactly
      53 bits. Rewriting
    </p>
    <pre><code>1 / 10 ~= J / (2**N)</code></pre>
    <p>as</p>
    <pre><code>J ~= 2**N / 10</code></pre>
    <p>
      and recalling that <em>J</em> has exactly 53 bits (is
      <code>&gt;= 2**52</code> but <code>&lt; 2**53</code>), the best value for
      <em>N</em> is 56:
    </p>
    <pre><code>&gt;&gt;&gt; 2**52 &lt;=  2**56 // 10  &lt; 2**53
True</code></pre>
    <p>
      That is, 56 is the only value for <em>N</em> that leaves <em>J</em> with
      exactly 53 bits. The best possible value for <em>J</em> is then that
      quotient rounded:
    </p>
    <pre><code>&gt;&gt;&gt; q, r = divmod(2**56, 10)
&gt;&gt;&gt; r
6</code></pre>
    <p>
      Since the remainder is more than half of 10, the best approximation is
      obtained by rounding up:
    </p>
    <pre><code>&gt;&gt;&gt; q+1
7205759403792794</code></pre>
    <p>
      Therefore the best possible approximation to 1/10 in 754 double precision
      is:
    </p>
    <pre><code>7205759403792794 / 2 ** 56</code></pre>
    <p>
      Dividing both the numerator and denominator by two reduces the fraction
      to:
    </p>
    <pre><code>3602879701896397 / 2 ** 55</code></pre>
    <p>
      Note that since we rounded up, this is actually a little bit larger than
      1/10; if we had not rounded up, the quotient would have been a little bit
      smaller than 1/10. But in no case can it be <em>exactly</em> 1/10!
    </p>
    <p>
      So the computer never “sees” 1/10: what it sees is the exact fraction
      given above, the best 754 double approximation it can get:
    </p>
    <pre><code>&gt;&gt;&gt; 0.1 * 2 ** 55
3602879701896397.0</code></pre>
    <p>
      If we multiply that fraction by 10**55, we can see the value out to 55
      decimal digits:
    </p>
    <pre><code>&gt;&gt;&gt; 3602879701896397 * 10 ** 55 // 2 ** 55
1000000000000000055511151231257827021181583404541015625</code></pre>
    <p>
      meaning that the exact number stored in the computer is equal to the
      decimal value 0.1000000000000000055511151231257827021181583404541015625.
      Instead of displaying the full decimal value, many languages (including
      older versions of Python), round the result to 17 significant digits:
    </p>
    <pre><code>&gt;&gt;&gt; format(0.1, &#39;.17f&#39;)
&#39;0.10000000000000001&#39;</code></pre>
    <p>
      The
      <a
        href="https://docs.python.org/3/library/fractions.html#module-fractions"
        class="reference internal"
        title="fractions: Rational numbers."
        ><code class="sourceCode python">fractions</code></a
      >
      and
      <a
        href="https://docs.python.org/3/library/decimal.html#module-decimal"
        class="reference internal"
        title="decimal: Implementation of the General Decimal Arithmetic  Specification."
        ><code class="sourceCode python">decimal</code></a
      >
      modules make these calculations easy:
    </p>
    <pre><code>&gt;&gt;&gt; from decimal import Decimal
&gt;&gt;&gt; from fractions import Fraction

&gt;&gt;&gt; Fraction.from_float(0.1)
Fraction(3602879701896397, 36028797018963968)

&gt;&gt;&gt; (0.1).as_integer_ratio()
(3602879701896397, 36028797018963968)

&gt;&gt;&gt; Decimal.from_float(0.1)
Decimal(&#39;0.1000000000000000055511151231257827021181583404541015625&#39;)

&gt;&gt;&gt; format(Decimal.from_float(0.1), &#39;.17&#39;)
&#39;0.10000000000000001&#39;</code></pre>
    <h3 id="table-of-contents">
      <a href="https://docs.python.org/3/contents.html">Table of Contents</a>
    </h3>
    <ul>
      <li>
        <a href="#" class="reference internal"
          >15. Floating Point Arithmetic: Issues and Limitations</a
        >
        <ul>
          <li>
            <a href="#representation-error" class="reference internal"
              >15.1. Representation Error</a
            >
          </li>
        </ul>
      </li>
    </ul>
    <h4 id="previous-topic">Previous topic</h4>
    <p>
      <a href="interactive.html" title="previous chapter"
        ><span class="section-number">14. </span>Interactive Input Editing and
        History Substitution</a
      >
    </p>
    <h4 id="next-topic">Next topic</h4>
    <p>
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